This is a question:
Prove that there are infinite natural numbers such that there is a latin square with size n & on the main & subsidiary diagonal numbers {1,2,...,n} appear.
I want all numbers with this property. I know for n=2,3,4 we can't make this latin square, But what about for numbers greater than five?
Are all numbers greater than five have this property?
consider a symmetrical matrix $A_{n\times n}$ we have $A_{ij} = A_{ji}$ we know that this doesn't violate the property of latin square, so it will be enough to consider upper or lower triangular matrix, this is because we would be using the same row or column only when $i=j$ which is $A_{ii}$ i.e an element of the diagonal.
Assume we have alredy filled $l$ rows let's count the number of ways we can fill the $l+1$ row, let us call the elements in this upper triangular row: $\alpha_1, \alpha_2, ...\alpha_m$, it is easy to see that $m = n - (l+1)$ (not considering an element of the diagonal).
for $\alpha_1$ we cannot use any element of the upper $l$ rows neither the element below this row (in the diagonal) neither the element at the right (in the diagonal), however from this vertical elements since the element $A_{l+1 l+1}$ is not included we would have a total of $n-(l+1)$ elements, by similar reasoning for element $\alpha_2$ we would have $n-(l+2)$, for $\alpha_m=n-(l+m) = 1$ so the number of ways for $\alpha_1, \alpha_2, ... \alpha_m = (n-(l+1))!$ it is easy to see that for any $l$ we would have a positive number of ways, being the worst case $l+1 = n-1$ then $(n-(n-2+1))! = 1$, this means that is possible to construct such a matrix.
From the construction of element $\alpha_1$ we can see that $n>2$